Prove that if $a, b, c$ are positive odd integers, then $b^2 - 4ac$ cannot be a perfect square.

If we let $a$, $b$, and $c$ be $2x+1$, $2y+1$, and $2z+1$ respectively, We will get a general from of a multiple of $8$ minus $3$. Since one portion of our from can have an $8$ factored out and the other portion be $4$ multiplied by the product of consecutive integers which is $4$ multiplied by an even number, giving us a multiple of $8$. So our expression is $-3 pmod 8$ or $5 pmod 8$, but note that any number squared is $0, 1, text { or } 4 pmod 8$. This is because any number is $0, 1, 2, 3, dots text { or } 7 pmod 8$. Squaring all those values $pmod 8$ will satisfy the statement. And so we are done.

$begin{align} b^2 - 4ac &= n^2 rightarrow text{ ...$n$ must be odd} \ (b - n)(b + n) &= 4k rightarrow text{ ...product of 2 even numbers} \ b-n=2 land b+n &=2k rightarrow text{ ...because $k$ is odd} \ 2n+2 &= 2k rightarrow\ n &= k - 1 text { ...contradicting $n$ and $k$ are odd.} end{align}$

$$a,b,c=text{odd}quadiffquad a=2A+1quad;quad b=2B+1quad;quad c=2C+1$$ $$Delta=b^2-4ac=n^2quadiffquad(2B+1)^2-4,(2A+1)(2C+1)=n^2$$ $$(4B^2+4B+1)-4,(4AC+2A+2C+1)=n^2$$ $$4,Big[(B^2+B)-(4AC+2A+2C+1)Big]+1=n^2$$ $$iff n=2k+1iff n^2=4k^2+4k+1iff$$ $$underbrace{underbrace{B(B+1)}_{even}-underbrace{2,(2AC+A+C)}_{even}-1}_{odd}=underbrace{k(k+1)}_{even}$$ Contradiction !

If $b^2-4ac$ was a perfect square then the polynomial $ax^2+bx+c$ would have some rationals $frac {p_1}{q_1}, frac {p_2}{q_2}$ as roots($frac{p_i}{q_i}=frac{-bpmsqrt{b^2-4ac}}{2a}$). Therefore $(q_1x-p_1)(q_2x-p_2)=ax^2+bx+c$.
So $q_1,q_2,p_1,p_2$ are odd integers (since $q_1q_2=a,p_1p_2=c$) and $q_1p_2+q_2p_1=-bRightarrowLeftarrow.$

Suppose $b^2-4ac$ is a square. Since $a,b,c$ are odd, $b^2-4ac$ must be an odd square. Hence $x=frac{-b+sqrt{b^2-4ac}}{2}$ is an integer. Since $x^2+bx+ac=0$, therefore $x^2+bx+acequiv 0,(mod,2)$. By FLT $x^2equiv x,(mod,2)$. Hence $(1+b)x+acequiv 0,(mod,2)$. Since $2|1+b$. Thus $0+acequiv 0,(mod,2)$ (contradiction)

HINT $$text{We know that all odd squares are of the form $8k+1$ (Why?)} tag{$star$}$$ Use $(star)$, to prove what you want. Move your mouse over the gray area for the complete solution.